Find the length x such that the two distances in the triangle are the same Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to position rectangles such that they are as close as possible to a reference point but do not overlap?Showing a line is parallel to a plane and finding the distance between themAn interesting point of a triangle. (Help needed to prove a statement.)How is a vertex of a triangle moving while another vertex is moving on its angle bisector?A conjecture about an angle on a solid bodyIs it possible to compute Right Triangle's Legs starting from another Right Triangle with the same Hypotenuse?Constructing a Regular Pentagon of a Desired LengthCalculate the projected distance on an inclined planeIs the blue area greater than the red area?In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.

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Find the length x such that the two distances in the triangle are the same
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to position rectangles such that they are as close as possible to a reference point but do not overlap?Showing a line is parallel to a plane and finding the distance between themAn interesting point of a triangle. (Help needed to prove a statement.)How is a vertex of a triangle moving while another vertex is moving on its angle bisector?A conjecture about an angle on a solid bodyIs it possible to compute Right Triangle's Legs starting from another Right Triangle with the same Hypotenuse?Constructing a Regular Pentagon of a Desired LengthCalculate the projected distance on an inclined planeIs the blue area greater than the red area?In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.
$begingroup$
I have been working on the following problem
Statement
Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).
Background
My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below
To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem
Solution
Using similar triangles one arrives at the three equations
$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$
Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.
Question
I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?
geometry triangles geometric-construction congruences-geometry
$endgroup$
add a comment |
$begingroup$
I have been working on the following problem
Statement
Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).
Background
My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below
To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem
Solution
Using similar triangles one arrives at the three equations
$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$
Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.
Question
I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?
geometry triangles geometric-construction congruences-geometry
$endgroup$
1
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
1
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago
add a comment |
$begingroup$
I have been working on the following problem
Statement
Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).
Background
My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below
To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem
Solution
Using similar triangles one arrives at the three equations
$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$
Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.
Question
I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?
geometry triangles geometric-construction congruences-geometry
$endgroup$
I have been working on the following problem
Statement
Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).
Background
My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below
To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem
Solution
Using similar triangles one arrives at the three equations
$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$
Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.
Question
I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?
geometry triangles geometric-construction congruences-geometry
geometry triangles geometric-construction congruences-geometry
edited 5 hours ago
N3buchadnezzar
asked 5 hours ago
N3buchadnezzarN3buchadnezzar
6,04233475
6,04233475
1
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
1
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago
add a comment |
1
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
1
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago
1
1
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
1
1
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.
Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.
Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.
Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.
By the exterior angle theorem, $mangle DEC = dfracalpha2$.
Hence $triangle EDC$ is isoceles.
So $CD = DE$.
$endgroup$
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
add a comment |
$begingroup$
The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.
Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.
Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.
$endgroup$
add a comment |
$begingroup$
Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$
$endgroup$
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
add a comment |
$begingroup$
Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$
Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.
Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.
Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.
Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.
By the exterior angle theorem, $mangle DEC = dfracalpha2$.
Hence $triangle EDC$ is isoceles.
So $CD = DE$.
$endgroup$
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
add a comment |
$begingroup$
Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.
Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.
Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.
Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.
By the exterior angle theorem, $mangle DEC = dfracalpha2$.
Hence $triangle EDC$ is isoceles.
So $CD = DE$.
$endgroup$
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
add a comment |
$begingroup$
Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.
Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.
Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.
Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.
By the exterior angle theorem, $mangle DEC = dfracalpha2$.
Hence $triangle EDC$ is isoceles.
So $CD = DE$.
$endgroup$
Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.
Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.
Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.
Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.
By the exterior angle theorem, $mangle DEC = dfracalpha2$.
Hence $triangle EDC$ is isoceles.
So $CD = DE$.
edited 43 mins ago
answered 4 hours ago
steven gregorysteven gregory
18.5k32359
18.5k32359
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
add a comment |
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
$begingroup$
This is exactly what I was looking for =)
$endgroup$
– N3buchadnezzar
3 hours ago
add a comment |
$begingroup$
The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.
Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.
Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.
$endgroup$
add a comment |
$begingroup$
The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.
Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.
Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.
$endgroup$
add a comment |
$begingroup$
The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.
Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.
Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.
$endgroup$
The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.
Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.
Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.
answered 4 hours ago
Michael BiroMichael Biro
11.7k21831
11.7k21831
add a comment |
add a comment |
$begingroup$
Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$
$endgroup$
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
add a comment |
$begingroup$
Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$
$endgroup$
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
add a comment |
$begingroup$
Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$
$endgroup$
Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$
edited 4 hours ago
answered 5 hours ago


Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
79.1k42867
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
add a comment |
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
$begingroup$
@mathmandan There was an edit after my comment.
$endgroup$
– Michael Biro
46 mins ago
add a comment |
$begingroup$
Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$
Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$
$endgroup$
add a comment |
$begingroup$
Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$
Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$
$endgroup$
add a comment |
$begingroup$
Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$
Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$
$endgroup$
Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$
Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$
edited 4 hours ago
answered 4 hours ago
VasyaVasya
4,5091618
4,5091618
add a comment |
add a comment |
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Lh joKGpYVQJ
1
$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
5 hours ago
$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
5 hours ago
$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
5 hours ago
1
$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
1 hour ago